7.1-bo'lim : Qismlar bo'yicha integratsiya
5. \( \displaystyle \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\) ni baholang.
Yechim
Bu erda birinchi qadam \(u\) va \(dv\) ni tanlashdir.
Bu holda biz eksponensialni \(u\) yoki \(dv\) ga, ikkinchisiga esa kosinusni qo'yishimiz mumkin. Bu tanlov juda muhim bo'lmagan bir nechta muammolardan biridir.
Ushbu muammo uchun \(u\) va \(dv\) uchun quyidagi variantlardan foydalaning.
\[u = \cos \left( {\frac{1}{4}z} \right)\hspace{0.5in}dv = {{\bf{e}}^{2z}}\,dz\]2-qadam
Keyinchalik, \(du\) (\(u\) farqlash orqali) va \(v\) (\(dv\) ni birlashtirish orqali) hisoblashimiz kerak.
\[\begin{align*}u & = \cos \left( {\frac{1}{4}z} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = - \frac{1}{4}\sin \left( {\frac{1}{4}z} \right)dz\\ dv & = \,{{\bf{e}}^{2z}}dz & \hspace{0.5in} & \to & \hspace{0.25in}v & = \frac{1}{2}{{\bf{e}}^{2z}}\end{align*}\]3-qadam
\(u\), \(du\), \(v\) va \(dv\) ni qismlar bo‘yicha integratsiyaga ulash formulasini beradi:
\[ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{8}\int{{{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right)\,dz}}\]4-qadam
Endi biz qismlar bo'yicha integratsiyani qayta bajarishimiz kerak va buning uchun biz quyidagi tanlovlardan foydalanamiz.
\[\begin{align*}u & = \sin \left( {\frac{1}{4}z} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = \frac{1}{4}\cos \left( {\frac{1}{4}z} \right)dz\\ dv & = \,{{\bf{e}}^{2z}}dz & \hspace{0.5in} & \to & \hspace{0.25in}v & = \frac{1}{2}{{\bf{e}}^{2z}}\end{align*}\]5-qadam
Bularni 3-bosqichdagi integralga ulash, beradi,
\[\begin{align*}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{8}\left[ {\frac{1}{2}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) - \frac{1}{8}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}} \right]\\ & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) - \frac{1}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\end{align*}\]6-qadam
Bu muammoni hal qilish uchun teng belgisining bir tomonida bir xil integrallarni olish uchun bir nechta asosiy algebraik manipulyatsiya qilishimiz kerak.
\[\begin{align*}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) - \frac{1}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\\ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} + \frac{1}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right)\\ \frac{{65}}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right)\end{align*}\]Nihoyat, biz qilishimiz kerak bo'lgan narsa - integraldagi koeffitsientni o'ng tomonga o'tkazish.
\[ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{{65}}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{4}{{65}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) + c}}\]